Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Proposition 2. Let \( G \) be a group and let \( a, b \in G \) . The equations \( {ax} = b \) and \( {ya} = b \) have unique solutions for \( x, y \in G \) . In particular, the left and right cancellation laws hold in \( G \), i.e.,\n\n(1) if \( {au} = {av} \), then \( u = v \), and\n\n(2) if \( {ub} = {vb} \), then \(... | Proof: We can solve \( {ax} = b \) by multiplying both sides on the left by \( {a}^{-1} \) and simplifying to get \( x = {a}^{-1}b \) . The uniqueness of \( x \) follows because \( {a}^{-1} \) is unique. Similarly, if \( {ya} = b, y = b{a}^{-1} \) . If \( {au} = {av} \), multiply both sides on the left by \( {a}^{-1} \... | Yes |
Proposition 1. (The Subgroup Criterion) A subset \( H \) of a group \( G \) is a subgroup if and only if\n\n(1) \( H \neq \varnothing \), and\n\n(2) for all \( x, y \in H, x{y}^{-1} \in H \) . | Proof: If \( H \) is a subgroup of \( G \), then certainly (1) and (2) hold because \( H \) contains the identity of \( G \) and the inverse of each of its elements and because \( H \) is closed under multiplication.\n\nIt remains to show conversely that if \( H \) satisfies both (1) and (2), then \( H \leq G \) . Let ... | Yes |
Proposition 2. If \( H = \langle x\rangle \), then \( \left| H\right| = \left| x\right| \) (where if one side of this equality is infinite, so is the other). More specifically\n\n(1) if \( \left| H\right| = n < \infty \), then \( {x}^{n} = 1 \) and \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are all the distinct elements of... | Proof: Let \( \left| x\right| = n \) and first consider the case when \( n < \infty \) . The elements \( 1, x,{x}^{2},\ldots ,{x}^{n - 1} \) are distinct because if \( {x}^{a} = {x}^{b} \), with, say, \( 0 \leq a < b < n \), then \( {x}^{b - a} = {x}^{0} = 1 \), contrary to \( n \) being the smallest positive power of ... | Yes |
Proposition 3. Let \( G \) be an arbitrary group, \( x \in G \) and let \( m, n \in \mathbb{Z} \) . If \( {x}^{n} = 1 \) and \( {x}^{m} = 1 \), then \( {x}^{d} = 1 \), where \( d = \left( {m, n}\right) \) . In particular, if \( {x}^{m} = 1 \) for some \( m \in \mathbb{Z} \), then \( \left| x\right| \) divides \( m \) . | Proof: By the Euclidean Algorithm (see Section 0.2 (6)) there exist integers \( r \) and \( s \) such that \( d = {mr} + {ns} \), where \( d \) is the g.c.d. of \( m \) and \( n \) . Thus\n\n\[ \n{x}^{d} = {x}^{{mr} + {ns}} = {\left( {x}^{m}\right) }^{r}{\left( {x}^{n}\right) }^{s} = {1}^{r}{1}^{s} = 1.\n\]\n\nThis pro... | Yes |
Theorem 4. Any two cyclic groups of the same order are isomorphic. More specifically, (1) if \( n \in {\mathbb{Z}}^{ + } \) and \( \langle x\rangle \) and \( \langle y\rangle \) are both cyclic groups of order \( n \), then the map\n\n\[ \varphi : \langle x\rangle \rightarrow \langle y\rangle \]\n\n\[ {x}^{k} \mapsto {... | Proof: Suppose \( \langle x\rangle \) and \( \langle y\rangle \) are both cyclic groups of order \( n \) . Let \( \varphi : \langle x\rangle \rightarrow \langle y\rangle \) be defined by \( \varphi \left( {x}^{k}\right) = {y}^{k} \) ; we must first prove \( \varphi \) is well defined, that is,\n\n\[ \text{if}{x}^{r} = ... | Yes |
Proposition 5. Let \( G \) be a group, let \( x \in G \) and let \( a \in \mathbb{Z} - \{ 0\} \) . (1) If \( \left| x\right| = \infty \), then \( \left| {x}^{a}\right| = \infty \) . | Proof: (1) By way of contradiction assume \( \left| x\right| = \infty \) but \( \left| {x}^{a}\right| = m < \infty \) . By definition of order \[ 1 = {\left( {x}^{a}\right) }^{m} = {x}^{am}. \] Also, \[ {x}^{-{am}} = {\left( {x}^{am}\right) }^{-1} = {1}^{-1} = 1. \] Now one of \( a{\;m} \) or \( - a{\;m} \) is positive... | Yes |
Proposition 6. Let \( H = \langle x\rangle \). (2) Assume \( \left| x\right| = n < \infty \). Then \( H = \left\langle {x}^{a}\right\rangle \) if and only if \( \left( {a, n}\right) = 1 \). In particular, the number of generators of \( H \) is \( \varphi \left( n\right) \) (where \( \varphi \) is Euler’s \( \varphi \) ... | Proof: We leave (1) as an exercise. In (2) if \( \left| x\right| = n < \infty \), Proposition 2 says \( {x}^{a} \) generates a subgroup of \( H \) of order \( \left| {x}^{a}\right| \). This subgroup equals all of \( H \) if and only if \( \left| {x}^{a}\right| = \left| x\right| \). By Proposition 5,\n\n\[ \left| {x}^{a... | No |
Theorem 7. Let \( H = \langle x\rangle \) be a cyclic group.\n\n(1) Every subgroup of \( H \) is cyclic. More precisely, if \( K \leq H \), then either \( K = \{ 1\} \) or \( K = \left\langle {x}^{d}\right\rangle \), where \( d \) is the smallest positive integer such that \( {x}^{d} \in K \) . | Proof: (1) Let \( K \leq H \) . If \( K = \{ 1\} \), the proposition is true for this subgroup, so we assume \( K \neq \{ 1\} \) . Thus there exists some \( a \neq 0 \) such that \( {x}^{a} \in K \) . If \( a < 0 \) then since \( K \) is a group also \( {x}^{-a} = {\left( {x}^{a}\right) }^{-1} \in K \) . Hence \( K \) ... | Yes |
Proposition 8. If \( \mathcal{A} \) is any nonempty collection of subgroups of \( G \), then the intersection of all members of \( \mathcal{A} \) is also a subgroup of \( G \) . | Proof: This is an easy application of the subgroup criterion (see also Exercise 10, Section 1). Let\n\n\[ K = \mathop{\bigcap }\limits_{{H \in \mathcal{A}}}H \]\n\nSince each \( H \in \mathcal{A} \) is a subgroup, \( 1 \in H \), so \( 1 \in K \), that is, \( K \neq \varnothing \). If \( a, b \in K \), then \( a, b \in ... | No |
Proposition 9. \( \bar{A} = \langle A\rangle \) . | Proof: We first prove \( \bar{A} \) is a subgroup. Note that \( \bar{A} \neq \varnothing \) (even if \( A = \varnothing \) ). If \( a, b \in \bar{A} \) with \( a = {a}_{1}^{{\epsilon }_{1}}{a}_{2}^{{\epsilon }_{2}}\ldots {a}_{n}^{{\epsilon }_{n}} \) and \( b = {b}_{1}^{{\delta }_{1}}{b}_{2}^{{\delta }_{2}}\ldots {b}_{m... | Yes |
Proposition 1. Let \( G \) and \( H \) be groups and let \( \varphi : G \rightarrow H \) be a homomorphism.\n\n(1) \( \varphi \left( {1}_{G}\right) = {1}_{H} \), where \( {1}_{G} \) and \( {1}_{H} \) are the identities of \( G \) and \( H \), respectively.\n\n(2) \( \varphi \left( {g}^{-1}\right) = \varphi {\left( g\ri... | Proof: (1) Since \( \varphi \left( {1}_{G}\right) = \varphi \left( {{1}_{G}{1}_{G}}\right) = \varphi \left( {1}_{G}\right) \varphi \left( {1}_{G}\right) \), the cancellation laws show that (1) holds.\n\n(2) \( \varphi \left( {1}_{G}\right) = \varphi \left( {g{g}^{-1}}\right) = \varphi \left( g\right) \varphi \left( {g}... | Yes |
Proposition 2. Let \( \varphi : G \rightarrow H \) be a homomorphism of groups with kernel \( K \) . Let\n\n\( X \in G/K \) be the fiber above \( a \), i.e., \( X = {\varphi }^{-1}\left( a\right) \) . Then\n\n(1) For any \( u \in X,\;X = \{ {uk} \mid k \in K\} \)\n\n(2) For any \( u \in X,\;X = \{ {ku} \mid k \in K\} \... | Proof: We prove (1) and leave the proof of (2) as an exercise. Let \( u \in X \) so, by definition of \( X,\varphi \left( u\right) = a \) . Let\n\n\[ \n{uK} = \{ {uk} \mid k \in K\} .\n\]\n\nWe first prove \( {uK} \subseteq X \) . For any \( k \in K \), \n\n\[ \n\varphi \left( {uk}\right) = \varphi \left( u\right) \var... | No |
Proposition 4. Let \( N \) be any subgroup of the group \( G \). The set of left cosets of \( N \) in \( G \) form a partition of \( G \). Furthermore, for all \( u, v \in G,{uN} = {vN} \) if and only if \( {v}^{-1}u \in N \) and in particular, \( {uN} = {vN} \) if and only if \( u \) and \( v \) are representatives of... | Proof: First of all note that since \( N \) is a subgroup of \( G,1 \in N \). Thus \( g = g \cdot 1 \in {gN} \) for all \( g \in G \), i.e.,\n\n\[ G = \mathop{\bigcup }\limits_{{g \in G}}{gN} \]\n\nTo show that distinct left cosets have empty intersection, suppose \( {uN} \cap {vN} \neq \varnothing \). We show \( {uN} ... | Yes |
Proposition 5. Let \( G \) be a group and let \( N \) be a subgroup of \( G \). (1) The operation on the set of left cosets of \( N \) in \( G \) described by \[ {uN} \cdot {vN} = \left( {uv}\right) N \] is well defined if and only if \( {gn}{g}^{-1} \in N \) for all \( g \in G \) and all \( n \in N \). (2) If the abov... | Proof: (1) Assume first that this operation is well defined, that is, for all \( u, v \in G \), \[ \text{if}u,{u}_{1} \in {uN}\text{and}v,{v}_{1} \in {vN}\;\text{then}\;{uvN} = {u}_{1}{v}_{1}N\text{.} \] Let \( g \) be an arbitrary element of \( G \) and let \( n \) be an arbitrary element of \( N \). Letting \( u = 1,... | Yes |
Theorem 6. Let \( N \) be a subgroup of the group \( G \) . The following are equivalent:\n\n(1) \( N \trianglelefteq G \)\n\n(2) \( {N}_{G}\left( N\right) = G \) (recall \( {N}_{G}\left( N\right) \) is the normalizer in \( G \) of \( N \) )\n\n(3) \( {gN} = {Ng} \) for all \( g \in G \)\n\n(4) the operation on left co... | Proof: We have already done the hard equivalences; the others are left as exercises. | No |
Proposition 7. A subgroup \( N \) of the group \( G \) is normal if and only if it is the kernel of some homomorphism. | Proof: If \( N \) is the kernel of the homomorphism \( \varphi \), then Proposition 2 shows that the left cosets of \( N \) are the same as the right cosets of \( N \) (and both are the fibers of the\n\nmap \( \varphi \) ). By (3) of Theorem 6, \( N \) is then a normal subgroup. (Another direct proof of this from the d... | No |
Theorem 8. (Lagrange’s Theorem) If \( G \) is a finite group and \( H \) is a subgroup of \( G \) , then the order of \( H \) divides the order of \( G \) (i.e., \( \left| H\right| \left| \right| G \mid \) ) and the number of left cosets of \( H \) in \( G \) equals \( \frac{\left| G\right| }{\left| H\right| } \) . | Proof: Let \( \left| H\right| = n \) and let the number of left cosets of \( H \) in \( G \) equal \( k \) . By\n\n# Proposition 4 the set of left cosets of \( H \) in \( G \) partition \( G \) . By definition of a left coset\n\nthe map:\n\n\[ H \rightarrow {gH}\;\text{ defined by }\;h \mapsto {gh} \]\n\nis a surjectio... | Yes |
Corollary 9. If \( G \) is a finite group and \( x \in G \), then the order of \( x \) divides the order of \( G \) . In particular \( {x}^{\left| G\right| } = 1 \) for all \( x \) in \( G \) . | Proof: By Proposition 2.2, \( \left| x\right| = \left| {\langle x\rangle }\right| \) . The first part of the corollary follows from Lagrange’s Theorem applied to \( H = \langle x\rangle \) . The second statement is clear since now \( \left| G\right| \) is a multiple of the order of \( x \) . | Yes |
Corollary 10. If \( G \) is a group of prime order \( p \), then \( G \) is cyclic, hence \( G \cong {Z}_{p} \) . | Proof: Let \( x \in G, x \neq 1 \) . Thus \( \left| {\langle x\rangle }\right| > 1 \) and \( \left| {\langle x\rangle }\right| \) divides \( \left| G\right| \) . Since \( \left| G\right| \) is prime we must have \( \left| {\langle x\rangle }\right| = \left| G\right| \), hence \( G = \langle x\rangle \) is cyclic (with ... | No |
Theorem 11. (Cauchy’s Theorem) If \( G \) is a finite group and \( p \) is a prime dividing \( \left| G\right| \) , then \( G \) has an element of order \( p \) . | Proof: We shall give a proof of this in the next chapter and another elegant proof is outlined in Exercise 9. | No |
Proposition 13. If \( H \) and \( K \) are finite subgroups of a group then\n\n\[ \left| {HK}\right| = \frac{\left| H\right| \left| K\right| }{\left| H \cap K\right| } \] | Proof: Notice that \( {HK} \) is a union of left cosets of \( K \), namely,\n\n\[ {HK} = \mathop{\bigcup }\limits_{{h \in H}}{hK} \]\n\nSince each coset of \( K \) has \( \left| K\right| \) elements it suffices to find the number of distinct left cosets of the form \( {hK}, h \in H \) . But \( {h}_{1}K = {h}_{2}K \) fo... | Yes |
Proposition 14. If \( H \) and \( K \) are subgroups of a group, \( {HK} \) is a subgroup if and only if \( {HK} = {KH} \) . | Proof: Assume first that \( {HK} = {KH} \) and let \( a, b \in {HK} \) . We prove \( a{b}^{-1} \in {HK} \) so \( {HK} \) is a subgroup by the subgroup criterion. Let\n\n\[ a = {h}_{1}{k}_{1}\;\text{ and }\;b = {h}_{2}{k}_{2}, \]\n\nfor some \( {h}_{1},{h}_{2} \in H \) and \( {k}_{1},{k}_{2} \in K \) . Thus \( {b}^{-1} ... | Yes |
Corollary 15. If \( H \) and \( K \) are subgroups of \( G \) and \( H \leq {N}_{G}\left( K\right) \), then \( {HK} \) is a subgroup of \( G \) . In particular, if \( K \trianglelefteq G \) then \( {HK} \leq G \) for any \( H \leq G \) . | Proof: We prove \( {HK} = {KH} \) . Let \( h \in H, k \in K \) . By assumption, \( {hk}{h}^{-1} \in K \) , hence\n\n\[ \n{hk} = \left( {{hk}{h}^{-1}}\right) h \in {KH}.\n\]\n\nThis proves \( {HK} \subseteq {KH} \) . Similarly, \( {kh} = h\left( {{h}^{-1}{kh}}\right) \in {HK} \), proving the reverse containment. The cor... | Yes |
Corollary 17. Let \( \varphi : G \rightarrow H \) be a homomorphism of groups.\n\n(1) \( \varphi \) is injective if and only if \( \ker \varphi = 1 \) . | Proof: Exercise. | No |
Theorem 18. (The Second or Diamond Isomorphism Theorem) Let \( G \) be a group, let \( A \) and \( B \) be subgroups of \( G \) and assume \( A \leq {N}_{G}\left( B\right) \) . Then \( {AB} \) is a subgroup of \( G \) , \( B \trianglelefteq {AB}, A \cap B \trianglelefteq A \) and \( {AB}/B \cong A/A \cap B \) . | Proof: By Corollary 15, \( {AB} \) is a subgroup of \( G \) . Since \( A \leq {N}_{G}\left( B\right) \) by assumption and \( B \leq {N}_{G}\left( B\right) \) trivially, it follows that \( {AB} \leq {N}_{G}\left( B\right) \), i.e., \( B \) is a normal subgroup of the subgroup \( {AB} \) .\n\nSince \( B \) is normal in \... | Yes |
Theorem 19. (The Third Isomorphism Theorem) Let \( G \) be a group and let \( H \) and \( K \) be normal subgroups of \( G \) with \( H \leq K \). Then \( K/H \trianglelefteq G/H \) and\n\n\[ \left( {G/H}\right) /\left( {K/H}\right) \cong G/K\text{.} \] | Proof: We leave as an easy exercise the verification that \( K/H \trianglelefteq G/H \). Define\n\n\[ \varphi : G/H \rightarrow G/K \]\n\n\[ \left( {gH}\right) \mapsto {gK}\text{.} \]\n\nTo show \( \varphi \) is well defined suppose \( {g}_{1}H = {g}_{2}H \). Then \( {g}_{1} = {g}_{2}h \), for some \( h \in H \). Becau... | No |
Theorem 20. (The Fourth or Lattice Isomorphism Theorem) Let \( G \) be a group and let \( N \) be a normal subgroup of \( G \) . Then there is a bijection from the set of subgroups \( A \) of \( G \) which contain \( N \) onto the set of subgroups \( \bar{A} = A/N \) of \( G/N \) . In particular, every subgroup of \( \... | Proof: The complete preimage of a subgroup in \( G/N \) is a subgroup of \( G \) by Exercise 1 of Section 1. The numerous details of the theorem to check are all completely straightforward. We therefore leave the proof of this theorem to the exercises. | No |
Proposition 21. If \( G \) is a finite abelian group and \( p \) is a prime dividing \( \left| G\right| \), then \( G \) contains an element of order \( p \) . | Proof: The proof proceeds by induction on \( \left| G\right| \), namely, we assume the result is valid for every group whose order is strictly smaller than the order of \( G \) and then prove the result valid for \( G \) (this is sometimes referred to as complete induction). Since \( \left| G\right| > 1 \), there is an... | Yes |
Theorem 22. (Jordan-Hölder) Let \( G \) be a finite group with \( G \neq 1 \) . Then\n\n(1) \( G \) has a composition series and\n\n(2) The composition factors in a composition series are unique, namely, if \( 1 = {N}_{0} \leq {N}_{1} \leq \cdots \leq {N}_{r} = G \) and \( 1 = {M}_{0} \leq {M}_{1} \leq \cdots \leq {M}_... | Proof: This is fairly straightforward. Since we shall not explicitly use this theorem to prove others in the text we outline the proof in a series of exercises at the end of this section. | No |
Proposition 23. The map \( \epsilon : {S}_{n} \rightarrow \{ \pm 1\} \) is a homomorphism (where \( \{ \pm 1\} \) is a multiplicative version of the cyclic group of order 2). | Proof: By definition,\n\n\[ \left( {\tau \sigma }\right) \left( \Delta \right) = \mathop{\prod }\limits_{{1 \leq i < j \leq n}}\left( {{x}_{{\tau \sigma }\left( i\right) } - {x}_{{\tau \sigma }\left( j\right) }}\right) \]\n\nSuppose that \( \sigma \left( \Delta \right) \) has exactly \( k \) factors of the form \( {x}_... | Yes |
Proposition 24. Transpositions are all odd permutations and \( \epsilon \) is a surjective homomorphism. | Moreover, since \( \epsilon \) is a homomorphism and every \( \sigma \in {S}_{n} \) is a product of transpositions, say \( \sigma = {\tau }_{1}{\tau }_{2}\cdots {\tau }_{k} \), then \( \epsilon \left( \sigma \right) = \epsilon \left( {\tau }_{1}\right) \cdots \epsilon \left( {\tau }_{k}\right) \) ; since \( \epsilon \l... | Yes |
Proposition 25. The permutation \( \sigma \) is odd if and only if the number of cycles of even length in its cycle decomposition is odd. | For example, \( \sigma = \left( {123456}\right) \left( {789}\right) \left( {1011}\right) \left( {12131415}\right) \left( {161718}\right) \) has 3 cycles of even length, so \( \epsilon \left( \sigma \right) = - 1 \) . On the other hand, \( \tau = \left( {1128104}\right) \left( {213}\right) \left( {5117}\right) \left( {6... | No |
Proposition 2. Let \( G \) be a group acting on the nonempty set \( A \) . The relation on \( A \) defined by\n\n\[ a \sim b\;\text{ if and only if }\;a = g \cdot b\text{ for some }g \in G \]\n\nis an equivalence relation. For each \( a \in A \), the number of elements in the equivalence class containing \( a \) is \( ... | Proof: We first prove \( \sim \) is an equivalence relation. By axiom 2 of an action, \( a = 1 \cdot a \) for all \( a \in A \), i.e., \( a \sim a \) and the relation is reflexive. If \( a \sim b \), then \( a = g \cdot b \) for some \( b \in G \) so that\n\n\[ {g}^{-1} \cdot a = {g}^{-1} \cdot \left( {g \cdot b}\right... | Yes |
Theorem 3. Let \( G \) be a group, let \( H \) be a subgroup of \( G \) and let \( G \) act by left multiplication on the set \( A \) of left cosets of \( H \) in \( G \) . Let \( {\pi }_{H} \) be the associated permutation representation afforded by this action. Then\n\n(1) \( G \) acts transitively on \( A \)\n\n(2) ... | Proof: To see that \( G \) acts transitively on \( A \), let \( {aH} \) and \( {bH} \) be any two elements of \( A \), and let \( g = b{a}^{-1} \) . Then \( g \cdot {aH} = \left( {b{a}^{-1}}\right) {aH} = {bH} \), and so the two arbitrary elements \( {aH} \) and \( {bH} \) of \( A \) lie in the same orbit, which proves... | Yes |
Corollary 4. (Cayley's Theorem) Every group is isomorphic to a subgroup of some symmetric group. If \( G \) is a group of order \( n \), then \( G \) is isomorphic to a subgroup of \( {S}_{n} \) . | Proof: Let \( H = 1 \) and apply the preceding theorem to obtain a homomorphism of \( G \) into \( {S}_{G} \) (here we are identifying the cosets of the identity subgroup with the elements of \( G \) ). Since the kernel of this homomorphism is contained in \( H = 1, G \) is isomorphic to its image in \( {S}_{G} \) . | No |
Corollary 5. If \( G \) is a finite group of order \( n \) and \( p \) is the smallest prime dividing \( \left| G\right| \) , then any subgroup of index \( p \) is normal. | Proof: Suppose \( H \leq G \) and \( \left| {G : H}\right| = p \) . Let \( {\pi }_{H} \) be the permutation representation afforded by multiplication on the set of left cosets of \( H \) in \( G \), let \( K = \ker {\pi }_{H} \) and let \( \left| {H : K}\right| = k \) . Then \( \left| {G : K}\right| = \left| {G : H}\ri... | Yes |
Proposition 6. The number of conjugates of a subset \( S \) in a group \( G \) is the index of the normalizer of \( S,\left| {G : {N}_{G}\left( S\right) }\right| \) . In particular, the number of conjugates of an element \( s \) of \( G \) is the index of the centralizer of \( s,\left| {G : {C}_{G}\left( s\right) }\rig... | Proof: The second assertion of the proposition follows from the observation that \( {N}_{G}\left( {\{ s\} }\right) = {C}_{G}\left( s\right) \) . | No |
Theorem 7. (The Class Equation) Let \( G \) be a finite group and let \( {g}_{1},{g}_{2},\ldots ,{g}_{r} \) be representatives of the distinct conjugacy classes of \( G \) not contained in the center \( Z\left( G\right) \) of \( G \) . Then\n\n\[ \left| G\right| = \left| {Z\left( G\right) }\right| + \mathop{\sum }\limi... | Proof: As noted in Example 2 above the element \( \{ x\} \) is a conjugacy class of size 1 if and only if \( x \in Z\left( G\right) \), since then \( {gx}{g}^{-1} = x \) for all \( g \in G \) . Let \( Z\left( G\right) = \left\{ {1,{z}_{2},\ldots ,{z}_{m}}\right\} \) , let \( {\mathcal{K}}_{1},{\mathcal{K}}_{2},\ldots ,... | Yes |
Theorem 8. If \( p \) is a prime and \( P \) is a group of prime power order \( {p}^{\alpha } \) for some \( \alpha \geq 1 \) , then \( P \) has a nontrivial center: \( Z\left( P\right) \neq 1 \) . | Proof: By the class equation\n\n\[ \left| P\right| = \left| {Z\left( P\right) }\right| + \mathop{\sum }\limits_{{i = 1}}^{r}\left| {P : {C}_{P}\left( {g}_{i}\right) }\right| \]\n\nwhere \( {g}_{1},\ldots ,{g}_{r} \) are representatives of the distinct non-central conjugacy classes. By definition, \( {C}_{P}\left( {g}_{... | Yes |
Corollary 9. If \( \\left| P\\right| = {p}^{2} \) for some prime \( p \), then \( P \) is abelian. More precisely, \( P \) is isomorphic to either \( {Z}_{{p}^{2}} \) or \( {Z}_{p} \\times {Z}_{p} \) . | Proof: Since \( Z\\left( P\\right) \\neq 1 \) by the theorem, it follows that \( P/Z\\left( P\\right) \) is cyclic. By Exercise 36, Section 3.1, \( P \) is abelian. If \( P \) has an element of order \( {p}^{2} \), then \( P \) is cyclic. Assume therefore that every nonidentity element of \( P \) has order \( p \) . Le... | Yes |
Proposition 10. Let \( \sigma ,\tau \) be elements of the symmetric group \( {S}_{n} \) and suppose \( \sigma \) has cycle decomposition\n\n\[ \left( {{a}_{1}{a}_{2}\ldots {a}_{{k}_{1}}}\right) \left( {{b}_{1}{b}_{2}\ldots {b}_{{k}_{2}}}\right) \ldots \]\n\nThen \( {\tau \sigma }{\tau }^{-1} \) has cycle decomposition\... | Proof: Observe that if \( \sigma \left( i\right) = j \), then\n\n\[ {\tau \sigma }{\tau }^{-1}\left( {\tau \left( i\right) }\right) = \tau \left( j\right) \]\n\nThus, if the ordered pair \( i, j \) appears in the cycle decomposition of \( \sigma \), then the ordered pair \( \tau \left( i\right) ,\tau \left( j\right) \)... | Yes |
Proposition 11. Two elements of \( {S}_{n} \) are conjugate in \( {S}_{n} \) if and only if they have the same cycle type. The number of conjugacy classes of \( {S}_{n} \) equals the number of partitions of \( n \) . | Proof: By Proposition 10, conjugate permutations have the same cycle type. Conversely, suppose the permutations \( {\sigma }_{1} \) and \( {\sigma }_{2} \) have the same cycle type. Order the cycles in nondecreasing length, including 1-cycles (if several cycles of \( {\sigma }_{1} \) and \( {\sigma }_{2} \) have the sa... | Yes |
Corollary 14. If \( K \) is any subgroup of the group \( G \) and \( g \in G \), then \( K \cong {gK}{g}^{-1} \) . Conjugate elements and conjugate subgroups have the same order. | Proof: Letting \( G = H \) in the proposition shows that conjugation by \( g \in G \) is an automorphism of \( G \), from which the corollary follows. | No |
For any subgroup \( H \) of a group \( G \), the quotient group \( {N}_{G}\left( H\right) /{C}_{G}\left( H\right) \) is isomorphic to a subgroup of \( \operatorname{Aut}\left( H\right) \) . In particular, \( G/Z\left( G\right) \) is isomorphic to a subgroup of \( \operatorname{Aut}\left( G\right) \) . | Proof: Since \( H \) is a normal subgroup of the group \( {N}_{G}\left( H\right) \), Proposition 13 (applied with \( {N}_{G}\left( H\right) \) playing the role of \( G \) ) implies the first assertion. The second assertion is the special case when \( H = G \), in which case \( {N}_{G}\left( G\right) = G \) and \( {C}_{... | Yes |
Proposition 16. The automorphism group of the cyclic group of order \( n \) is isomorphic to \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ \times } \), an abelian group of order \( \varphi \left( n\right) \) (where \( \varphi \) is Euler’s function). | Proof: Let \( x \) be a generator of the cyclic group \( {Z}_{n} \) . If \( \psi \in \operatorname{Aut}\left( {Z}_{n}\right) \), then \( \psi \left( x\right) = {x}^{a} \) for some \( a \in \mathbb{Z} \) and the integer \( a \) uniquely determines \( \psi \) . Denote this automorphism by \( {\psi }_{a} \) . As usual, si... | Yes |
Lemma 19. Let \( P \in {Sy}{l}_{p}\left( G\right) \) . If \( Q \) is any \( p \) -subgroup of \( G \), then \( Q \cap {N}_{G}\left( P\right) = Q \cap P \) . | Proof: Let \( H = {N}_{G}\left( P\right) \cap Q \) . Since \( P \leq {N}_{G}\left( P\right) \) it is clear that \( P \cap Q \leq H \), so we must prove the reverse inclusion. Since by definition \( H \leq Q \), this is equivalent to showing \( H \leq P \) . We do this by demonstrating that \( {PH} \) is a \( p \) -subg... | Yes |
Corollary 20. Let \( P \) be a Sylow \( p \) -subgroup of \( G \) . Then the following are equivalent:\n\n(1) \( P \) is the unique Sylow \( p \) -subgroup of \( G \), i.e., \( {n}_{p} = 1 \)\n\n(2) \( P \) is normal in \( G \)\n\n(3) \( P \) is characteristic in \( G \)\n\n(4) All subgroups generated by elements of \(... | Proof: If (1) holds, then \( {gP}{g}^{-1} = P \) for all \( g \in G \) since \( {gP}{g}^{-1} \in {Sy}{l}_{p}\left( G\right) \), i.e., \( P \) is normal in \( G \) . Hence (1) implies (2). Conversely, if \( P \trianglelefteq G \) and \( Q \in {\operatorname{Syl}}_{p}\left( G\right) \), then by Sylow’s Theorem there exis... | Yes |
Proposition 21. If \( \left| G\right| = {60} \) and \( G \) has more than one Sylow 5-subgroup, then \( G \) is simple. | Proof: Suppose by way of contradiction that \( \left| G\right| = {60} \) and \( {n}_{5} > 1 \) but that there exists \( H \) a normal subgroup of \( G \) with \( H \neq 1 \) or \( G \) . By Sylow’s Theorem the only possibility for \( {n}_{5} \) is 6 . Let \( P \in {\operatorname{Syl}}_{5}\left( G\right) \), so that \( ... | Yes |
Proposition 23. If \( G \) is a simple group of order 60, then \( G \cong {A}_{5} \) . | Proof: Let \( G \) be a simple group of order 60, so \( {n}_{2} = 3,5 \) or 15 . Let \( P \in {\operatorname{Syl}}_{2}\left( G\right) \) and let \( N = {N}_{G}\left( P\right) \), so \( \left| {G : N}\right| = {n}_{2} \) . First observe that \( G \) has no proper subgroup \( H \) of index less that 5, as follows: if \( ... | Yes |
Proposition 1. If \( {G}_{1},\ldots ,{G}_{n} \) are groups, their direct product is a group of order \( \left| {G}_{1}\right| \left| {G}_{2}\right| \cdots \left| {G}_{n}\right| \) (if any \( {G}_{i} \) is infinite, so is the direct product). | Proof: Let \( G = {G}_{1} \times {G}_{2} \times \cdots \times {G}_{n} \) . The proof that the group axioms hold for \( G \) is straightforward since each axiom is a consequence of the fact that the same axiom holds in each factor, \( {G}_{i} \), and the operation on \( G \) is defined componentwise. For example, the as... | Yes |
For each fixed \( i \) the set of elements of \( G \) which have the identity of \( {G}_{j} \) in the \( {j}^{\text{th }} \) position for all \( j \neq i \) and arbitrary elements of \( {G}_{i} \) in position \( i \) is a subgroup of \( G \) isomorphic to \( {G}_{i} \): | Since the operation in \( G \) is defined componentwise, it follows easily from the subgroup criterion that \( \left\{ {\left( {1,1,\ldots ,1,{g}_{i},1,\ldots ,1}\right) \mid {g}_{i} \in {G}_{i}}\right\} \) is a subgroup of \( G \) . Furthermore, the map \( {g}_{i} \mapsto \left( {1,1,\ldots ,1,{g}_{i},1,\ldots ,1}\rig... | Yes |
Theorem 3. (Fundamental Theorem of Finitely Generated Abelian Groups) Let \( G \) be a finitely generated abelian group. Then (1)\n\n\[ G \cong {\mathbb{Z}}^{r} \times {Z}_{{n}_{1}} \times {Z}_{{n}_{2}} \times \cdots \times {Z}_{{n}_{s}} \]\n\nfor some integers \( r,{n}_{1},{n}_{2},\ldots ,{n}_{s} \) satisfying the fol... | Proof: We shall derive this theorem in Section 12.1 as a consequence of a more general classification theorem. For finite groups we shall give an alternate proof at the end of Section 6.1. | No |
Proposition 6. Let \( m, n \in {\mathbb{Z}}^{ + } \). (1) \( {Z}_{m} \times {Z}_{n} \cong {Z}_{mn} \) if and only if \( \left( {m, n}\right) = 1 \). | Proof: Since (2) is an easy exercise using (1) and induction on \( k \), we concentrate on proving (1). Let \( {Z}_{m} = \langle x\rangle ,{Z}_{n} = \langle y\rangle \) and let \( l = \) l.c.m. \( \left( {m, n}\right) \) . Note that \( l = {mn} \) if and only if \( \left( {m, n}\right) = 1 \) . Let \( {x}^{a}{y}^{b} \)... | No |
Proposition 7. Let \( G \) be a group, let \( x, y \in G \) and let \( H \leq G \) . Then\n\n(1) \( {xy} = {yx}\left\lbrack {x, y}\right\rbrack \) (in particular, \( {xy} = {yx} \) if and only if \( \left\lbrack {x, y}\right\rbrack = 1 \) ). | Proof: (1) This is immediate from the definition of \( \left\lbrack {x, y}\right\rbrack \) . | No |
Proposition 8. Let \( H \) and \( K \) be subgroups of the group \( G \). The number of distinct ways of writing each element of the set \( {HK} \) in the form \( {hk} \), for some \( h \in H \) and \( k \in K \) is \( \left| {H \cap K}\right| \). In particular, if \( H \cap K = 1 \), then each element of \( {HK} \) ca... | Proof: Exercise. | No |
Theorem 9. Suppose \( G \) is a group with subgroups \( H \) and \( K \) such that\n\n(1) \( H \) and \( K \) are normal in \( G \), and\n\n(2) \( H \cap K = 1 \).\n\nThen \( {HK} \cong H \times K \) . | Proof: Observe that by hypothesis (1), \( {HK} \) is a subgroup of \( G \) (see Corollary 3.15). Let \( h \in H \) and let \( k \in K \) . Since \( H \trianglelefteq G,{k}^{-1}{hk} \in H \), so that \( {h}^{-1}\left( {{k}^{-1}{hk}}\right) \in H \) . Similarly, \( \left( {{h}^{-1}{k}^{-1}h}\right) k \in K \) . Since \( ... | Yes |
Theorem 10. Let \( H \) and \( K \) be groups and let \( \varphi \) be a homomorphism from \( K \) into Aut \( \left( H\right) \) . Let \( \cdot \) denote the (left) action of \( K \) on \( H \) determined by \( \varphi \) . Let \( G \) be the set of ordered pairs \( \left( {h, k}\right) \) with \( h \in H \) and \( k ... | Proof: It is straightforward to check that \( G \) is a group under this multiplication using the fact that \( \cdot \) is an action of \( K \) on \( H \) . For example, the associative law is verified as follows:\n\n\[ \left( {\left( {a, x}\right) \left( {b, y}\right) }\right) \left( {c, z}\right) = \left( {{ax} \cdot... | No |
Proposition 11. Let \( H \) and \( K \) be groups and let \( \varphi : K \rightarrow \operatorname{Aut}\left( H\right) \) be a homomorphism. Then the following are equivalent:\n\n(1) the identity (set) map between \( H \rtimes K \) and \( H \times K \) is a group homomorphism (hence an isomorphism)\n\n(2) \( \varphi \)... | Proof: \( \left( 1\right) \Rightarrow \left( 2\right) \) By definition of the group operation in \( H \rtimes K \)\n\n\[ \left( {{h}_{1},{k}_{1}}\right) \left( {{h}_{2},{k}_{2}}\right) = \left( {{h}_{1}{k}_{1} \cdot {h}_{2},{k}_{1}{k}_{2}}\right) \]\n\nfor all \( {h}_{1},{h}_{2} \in H \) and \( {k}_{1},{k}_{2} \in K \)... | Yes |
Theorem 12. Suppose \( G \) is a group with subgroups \( H \) and \( K \) such that\n\n(1) \( H \trianglelefteq G \), and\n\n(2) \( H \cap K = 1 \) .\n\nLet \( \varphi : K \rightarrow \operatorname{Aut}\left( H\right) \) be the homomorphism defined by mapping \( k \in K \) to the automorphism of left conjugation by \( ... | Proof: Note that since \( H \trianglelefteq G,{HK} \) is a subgroup of \( G \) . By Proposition 8 every element of \( {HK} \) can be written uniquely in the form \( {hk} \), for some \( h \in H \) and \( k \in K \) . Thus the map \( {hk} \mapsto \left( {h, k}\right) \) is a set bijection from \( {HK} \) onto \( H \rtim... | Yes |
Let \( p \) be a prime and let \( P \) be a group of order \( {p}^{a}, a \geq 1 \) . Then\n\n(1) The center of \( P \) is nontrivial: \( Z\left( P\right) \neq 1 \) .\n\n(2) If \( H \) is a nontrivial normal subgroup of \( P \) then \( H \) intersects the center non-trivially: \( H \cap Z\left( P\right) \neq 1 \) . In p... | These results rely ultimately on the class equation and it may be useful for the reader to review Section 4.3.\n\nPart 1 is Theorem 8 of Chapter 4 and is also the special case of part 2 when \( H = P \) . We therefore begin by proving (2); we shall not quote Theorem 8 of Chapter 4 although the argument that follows is ... | Yes |
Proposition 2. Let \( p \) be a prime and let \( P \) be a group of order \( {p}^{a} \) . Then \( P \) is nilpotent of nilpotence class at most \( a - 1 \) . | Proof: For each \( i \geq 0, P/{Z}_{i}\left( P\right) \) is a \( p \) -group, so\n\n\[ \text{if}\left| {P/{Z}_{i}\left( P\right) }\right| > 1\text{then}Z\left( {P/{Z}_{i}\left( P\right) }\right) \neq 1 \]\n\nby Theorem 1(1). Thus if \( {Z}_{i}\left( P\right) \neq G \) then \( \left| {{Z}_{i + 1}\left( P\right) }\right|... | Yes |
Theorem 3. Let \( G \) be a finite group, let \( {p}_{1},{p}_{2},\ldots ,{p}_{s} \) be the distinct primes dividing its order and let \( {P}_{i} \in {\operatorname{Syl}}_{{p}_{i}}\left( G\right) ,1 \leq i \leq s \) . Then the following are equivalent:\n\n(1) \( G \) is nilpotent\n\n(2) if \( H < G \) then \( H < {N}_{G... | Proof: The proof that (1) implies (2) is the same argument as for \( p \) -groups - the only fact we needed was if \( G \) is nilpotent then so is \( G/Z\left( G\right) \) - so the details are omitted (cf. the exercises).\n\nTo show that (2) implies (3) let \( P = {P}_{i} \) for some \( i \) and let \( N = {N}_{G}\left... | No |
Proposition 5. If \( G \) is a finite group such that for all positive integers \( n \) dividing its order, \( G \) contains at most \( n \) elements \( x \) satisfying \( {x}^{n} = 1 \), then \( G \) is cyclic. | Proof: Let \( \left| G\right| = {p}_{1}^{{\alpha }_{1}}\cdots {p}_{s}^{{\alpha }_{s}} \) and let \( {P}_{i} \) be a Sylow \( {p}_{i} \) -subgroup of \( G \) for \( i = 1,2,\ldots, s \) . Since \( {p}_{i}^{{\alpha }_{i}}\left| \right| G| \) and the \( {p}_{i}^{{\alpha }_{i}} \) elements of \( {P}_{i} \) are solutions of... | Yes |
Proposition 6. (Frattini’s Argument) Let \( G \) be a finite group, let \( H \) be a normal subgroup of \( G \) and let \( P \) be a Sylow \( p \) -subgroup of \( H \) . Then \( G = H{N}_{G}\left( P\right) \) and \( \left| {G : H}\right| \) divides \( \left| {{N}_{G}\left( P\right) }\right| \) . | Proof: By Corollary 3.15, \( H{N}_{G}\left( P\right) \) is a subgroup of \( G \) and \( H{N}_{G}\left( P\right) = {N}_{G}\left( P\right) H \) since \( H \) is a normal subgroup of \( G \) . Let \( g \in G \) . Since \( {P}^{g} \leq {H}^{g} = H \), both \( P \) and \( {P}^{g} \) are Sylow \( p \) -subgroups of \( H \) .... | Yes |
Proposition 7. A finite group is nilpotent if and only if every maximal subgroup is normal. | Proof: Let \( G \) be a finite nilpotent group and let \( M \) be a maximal subgroup of \( G \) . As in the proof of Theorem 1, since \( M < {N}_{G}\left( M\right) \) (by Theorem 3(2)) maximality of \( M \) forces \( {N}_{G}\left( M\right) = G \), i.e., \( M \trianglelefteq G \) .\n\nConversely, assume every maximal su... | Yes |
Theorem 8. A group \( G \) is nilpotent if and only if \( {G}^{n} = 1 \) for some \( n \geq 0 \) . More precisely, \( G \) is nilpotent of class \( c \) if and only if \( c \) is the smallest nonnegative integer such that \( {G}^{c} = 1 \) . If \( G \) is nilpotent of class \( c \) then\n\n\[ \n{Z}_{i}\left( G\right) \... | Proof: This is proved by a straightforward induction on the length of either the upper or lower central series. | No |
Theorem 9. A group \( G \) is solvable if and only if \( {G}^{\left( n\right) } = 1 \) for some \( n \geq 0 \) . | Proof: Assume first that \( G \) is solvable and so possesses a series\n\n\[ 1 = {H}_{0} \trianglelefteq {H}_{1} \trianglelefteq \cdots \trianglelefteq {H}_{s} = G \]\n\nsuch that each factor \( {H}_{i + 1}/{H}_{i} \) is abelian. We prove by induction that \( {G}^{\left( i\right) } \leq {H}_{s - i} \) . This is true fo... | Yes |
Proposition 10. Let \( G \) and \( K \) be groups, let \( H \) be a subgroup of \( G \) and let \( \varphi : G \rightarrow K \) be a surjective homomorphism.\n\n(1) \( {H}^{\left( i\right) } \leq {G}^{\left( i\right) } \) for all \( i \geq 0 \) . In particular, if \( G \) is solvable, then so is \( H \), i.e., subgroup... | Proof: Part 1 follows from the observation that since \( H \leq G \), by definition of commutator subgroups, \( \left\lbrack {H, H}\right\rbrack \leq \left\lbrack {G, G}\right\rbrack \), i.e., \( {H}^{\left( 1\right) } \leq {G}^{\left( 1\right) } \) . Then, by induction,\n\n\[ \n{H}^{\left( i\right) } \leq {G}^{\left( ... | Yes |
Theorem 11. Let \( G \) be a finite group.\n\n(1) (Burnside) If \( \left| G\right| = {p}^{a}{q}^{b} \) for some primes \( p \) and \( q \), then \( G \) is solvable. | We shall prove Burnside’s Theorem in Chapter 19 and deduce Philip Hall’s generalization of it. | No |
Lemma 13. In a finite group \( G \) if \( {n}_{p} ≢ 1\left( {\;\operatorname{mod}\;{p}^{2}}\right) \), then there are distinct Sylow \( p \) -subgroups \( P \) and \( R \) of \( G \) such that \( P \cap R \) is of index \( p \) in both \( P \) and \( R \) (hence is normal in each). | Proof: The argument is an easy refinement of the proof of the congruence part of Sylow’s Theorem (cf. the exercises at the end of Section 4.5). Let \( P \) act by conjugation on the set \( {\operatorname{Syl}}_{p}\left( G\right) \) . Let \( {\mathcal{O}}_{1},\ldots ,{\mathcal{O}}_{s} \) be the orbits under this action ... | Yes |
Theorem 17. Let \( G \) be a group, \( S \) a set and \( \varphi : S \rightarrow G \) a set map. Then there is a unique group homomorphism \( \Phi : F\left( S\right) \rightarrow G \) such that the following diagram commutes: | \( \textit{Proof: Such a map }\Phi \) must satisfy \( \Phi \left( {{s}_{1}^{{\epsilon }_{1}}{s}_{2}^{{\epsilon }_{2}}\ldots {s}_{n}^{{\epsilon }_{n}}}\right) = \varphi {\left( {s}_{1}\right) }^{{\epsilon }_{1}}\varphi {\left( {s}_{2}\right) }^{{\epsilon }_{2}}\ldots \varphi {\left( {s}_{n}\right) }^{{\epsilon }_{n}} \)... | Yes |
Corollary 18. \( F\\left( S\\right) \) is unique up to a unique isomorphism which is the identity map on the set \( S \) . | Proof: This follows from the universal property. Suppose \( F\\left( S\\right) \) and \( {F}^{\\prime }\\left( S\\right) \) are two free groups generated by \( S \) . Since \( S \) is contained in both \( F\\left( S\\right) \) and \( {F}^{\\prime }\\left( S\\right) \), we have natural injections \( S \\hookrightarrow {... | Yes |
Proposition 1. Let \( R \) be a ring. Then\n\n(1) \( {0a} = {a0} = 0 \) for all \( a \in R \) . | Proof: These all follow from the distributive laws and cancellation in the additive group \( R \) . For example,(1) follows from \( {0a} = \left( {0 + 0}\right) a = {0a} + {0a} \) . | No |
Proposition 2. Assume \( a, b \) and \( c \) are elements of any ring with \( a \) not a zero divisor. If \( {ab} = {ac} \), then either \( a = 0 \) or \( b = c \) (i.e., if \( a \neq 0 \) we can cancel the \( a \) ’s). In particular, if \( a, b, c \) are any elements in an integral domain and \( {ab} = {ac} \), then e... | Proof: If \( {ab} = {ac} \) then \( a\left( {b - c}\right) = 0 \) so either \( a = 0 \) or \( b - c = 0 \) . The second statement follows from the first and the definition of an integral domain. | Yes |
Corollary 3. Any finite integral domain is a field. | Proof: Let \( R \) be a finite integral domain and let \( a \) be a nonzero element of \( R \) . By the cancellation law the map \( x \mapsto {ax} \) is an injective function. Since \( R \) is finite this map is also surjective. In particular, there is some \( b \in R \) such that \( {ab} = 1 \), i.e., \( a \) is a uni... | Yes |
Proposition 4. Let \( R \) be an integral domain and let \( p\left( x\right), q\left( x\right) \) be nonzero elements of \( R\left\lbrack x\right\rbrack \) . Then\n\n(1) degree \( p\left( x\right) q\left( x\right) = \) degree \( p\left( x\right) + \) degree \( q\left( x\right) \),\n\n(2) the units of \( R\left\lbrack x... | Proof: If \( R \) has no zero divisors then neither does \( R\left\lbrack x\right\rbrack \) ; if \( p\left( x\right) \) and \( q\left( x\right) \) are polynomials with leading terms \( {a}_{n}{x}^{n} \) and \( {b}_{m}{x}^{m} \), respectively, then the leading term of \( p\left( x\right) q\left( x\right) \) is \( {a}_{n... | Yes |
(1) The image of \( \varphi \) is a subring of \( S \) . | Proof: (1) If \( {s}_{1},{s}_{2} \in \operatorname{im}\varphi \) then \( {s}_{1} = \varphi \left( {r}_{1}\right) \) and \( {s}_{2} = \varphi \left( {r}_{2}\right) \) for some \( {r}_{1},{r}_{2} \in R \) . Then \( \varphi \left( {{r}_{1} - {r}_{2}}\right) = {s}_{1} - {s}_{2} \) and \( \varphi \left( {{r}_{1}{r}_{2}}\rig... | Yes |
(1) (The First Isomorphism Theorem for Rings) If \( \varphi : R \rightarrow S \) is a homomorphism of rings, then the kernel of \( \varphi \) is an ideal of \( R \), the image of \( \varphi \) is a subring of \( S \) and \( R/\ker \varphi \) is isomorphic as a ring to \( \varphi \left( R\right) \) . | Proof: This is just a matter of collecting previous calculations. If \( I \) is the kernel of \( \varphi \), then the cosets (under addition) of \( I \) are precisely the fibers of \( \varphi \) . In particular, the cosets \( r + I, s + I \) and \( {rs} + I \) are the fibers of \( \varphi \) over \( \varphi \left( r\ri... | Yes |
Proposition 9. Let \( I \) be an ideal of \( R \). (1) \( I = R \) if and only if \( I \) contains a unit. | Proof: (1) If \( I = R \) then \( I \) contains the unit 1. Conversely, if \( u \) is a unit in \( I \) with inverse \( v \), then for any \( r \in R \)\n\n\[ r = r \cdot 1 = r\left( {vu}\right) = \left( {rv}\right) u \in I \]\n\nhence \( R = I \). | Yes |
Corollary 10. If \( R \) is a field then any nonzero ring homomorphism from \( R \) into another ring is an injection. | Proof: The kernel of a ring homomorphism is an ideal. The kernel of a nonzero homomorphism is a proper ideal hence is 0 by the proposition. | Yes |
Proposition 11. In a ring with identity every proper ideal is contained in a maximal ideal. | Proof: Let \( R \) be a ring with identity and let \( I \) be a proper ideal (so \( R \) cannot be the zero ring, i.e., \( 1 \neq 0 \) ). Let \( \mathcal{S} \) be the set of all proper ideals of \( R \) which contain \( I \) . Then \( \mathcal{S} \) is nonempty \( \left( {I \in \mathcal{S}}\right) \) and is partially o... | Yes |
Proposition 12. Assume \( R \) is commutative. The ideal \( M \) is a maximal ideal if and only if the quotient ring \( R/M \) is a field. | Proof: This follows from the Lattice Isomorphism Theorem together with Proposition 9(2). The ideal \( M \) is maximal if and only if there are no ideals \( I \) with \( M \subset I \subset R \) . By the Lattice Isomorphism Theorem the ideals of \( R \) containing \( M \) correspond bijectively with the ideals of \( R/M... | Yes |
Proposition 13. Assume \( R \) is commutative. Then the ideal \( P \) is a prime ideal in \( R \) if and only if the quotient ring \( R/P \) is an integral domain. | Proof: This proof is simply a matter of translating the definition of a prime ideal into the language of quotients. The ideal \( P \) is prime if and only if \( P \neq R \) and whenever \( {ab} \in P \), then either \( a \in P \) or \( b \in P \) . Use the bar notation for elements of \( R/P \) : \( \bar{r} = r + P \) ... | Yes |
Corollary 14. Assume \( R \) is commutative. Every maximal ideal of \( R \) is a prime ideal. | Proof: If \( M \) is a maximal ideal then \( R/M \) is a field by Proposition 12. A field is an integral domain so the corollary follows from Proposition 13. | Yes |
Corollary 16. Let \( R \) be an integral domain and let \( Q \) be the field of fractions of \( R \) . If a field \( F \) contains a subring \( {R}^{\prime } \) isomorphic to \( R \) then the subfield of \( F \) generated by \( {R}^{\prime } \) is isomorphic to \( Q \) . | Proof: Let \( \varphi : R \cong {R}^{\prime } \subseteq F \) be a (ring) isomorphism of \( R \) to \( {R}^{\prime } \) . In particular, \( \varphi : R \rightarrow F \) is an injective homomorphism from \( R \) into the field \( F \) . Let \( \Phi : Q \rightarrow F \) be the extension of \( \varphi \) to \( Q \) as in t... | Yes |
Theorem 17. (Chinese Remainder Theorem) Let \( {A}_{1},{A}_{2},\ldots ,{A}_{k} \) be ideals in \( R \) . The map \( R \rightarrow R/{A}_{1} \times R/{A}_{2} \times \cdots \times R/{A}_{k}\; \) defined by \( \;r \mapsto \left( {r + {A}_{1}, r + {A}_{2},\ldots, r + {A}_{k}}\right) \) is a ring homomorphism with kernel \(... | Proof: We first prove this for \( k = 2 \) ; the general case will follow by induction. Let \( A = {A}_{1} \) and \( B = {A}_{2} \) . Consider the map \( \varphi : R \rightarrow R/A \times R/B \) defined by \( \varphi \left( r\right) = \left( {r{\;\operatorname{mod}\;A}, r{\;\operatorname{mod}\;B}}\right) \), where \( ... | Yes |
Corollary 18. Let \( n \) be a positive integer and let \( {p}_{1}{}^{{\alpha }_{1}}{p}_{2}{}^{{\alpha }_{2}}\ldots {p}_{k}{}^{{\alpha }_{k}} \) be its factorization into powers of distinct primes. Then\n\n\[ \n\mathbb{Z}/n\mathbb{Z} \cong \left( {\mathbb{Z}/{p}_{1}{}^{{\alpha }_{1}}\mathbb{Z}}\right) \times \left( {\m... | If we compare orders on the two sides of this last isomorphism, we obtain the formula\n\n\[ \n\varphi \left( n\right) = \varphi \left( {{p}_{1}{}^{{\alpha }_{1}}}\right) \varphi \left( {{p}_{2}{}^{{\alpha }_{2}}}\right) \ldots \varphi \left( {{p}_{k}{}^{{\alpha }_{k}}}\right)\n\]\n\nfor the Euler \( \varphi \) -functio... | Yes |
Proposition 1. Every ideal in a Euclidean Domain is principal. More precisely, if \( I \) is any nonzero ideal in the Euclidean Domain \( R \) then \( I = \left( d\right) \), where \( d \) is any nonzero element of \( I \) of minimum norm. | Proof: If \( I \) is the zero ideal, there is nothing to prove. Otherwise let \( d \) be any nonzero element of \( I \) of minimum norm (such a \( d \) exists since the set \( \{ N\left( a\right) \mid a \in I\} \) has a minimum element by the Well Ordering of \( \mathbb{Z} \) ). Clearly \( \left( d\right) \subseteq I \... | Yes |
Proposition 3. Let \( R \) be an integral domain. If two elements \( d \) and \( {d}^{\prime } \) of \( R \) generate the same principal ideal, i.e., \( \left( d\right) = \left( {d}^{\prime }\right) \), then \( {d}^{\prime } = {ud} \) for some unit \( u \) in \( R \) . In particular, if \( d \) and \( {d}^{\prime } \) ... | Proof: This is clear if either \( d \) or \( {d}^{\prime } \) is zero so we may assume \( d \) and \( {d}^{\prime } \) are nonzero. Since \( d \in \left( {d}^{\prime }\right) \) there is some \( x \in R \) such that \( d = x{d}^{\prime } \) . Since \( {d}^{\prime } \in \left( d\right) \) there is some \( y \in R \) suc... | Yes |
Theorem 4. Let \( R \) be a Euclidean Domain and let \( a \) and \( b \) be nonzero elements of \( R \). Let \( d = {r}_{n} \) be the last nonzero remainder in the Euclidean Algorithm for \( a \) and \( b \) described at the beginning of this chapter. Then\n\n(1) \( d \) is a greatest common divisor of \( a \) and \( b... | Proof: By Proposition 1, the ideal generated by \( a \) and \( b \) is principal so \( a, b \) do have a greatest common divisor, namely any element which generates the (principal) ideal \( \left( {a, b}\right) \). Both parts of the theorem will follow therefore once we show \( d = {r}_{n} \) generates this ideal, i.e.... | Yes |
Proposition 5. Let \( R \) be an integral domain that is not a field. If \( R \) is a Euclidean Domain then there are universal side divisors in \( R \) . | Proof: Suppose \( R \) is Euclidean with respect to some norm \( N \) and let \( u \) be an element of \( R - \widetilde{R} \) (which is nonempty since \( R \) is not a field) of minimal norm. For any \( x \in R \) , write \( x = {qu} + r \) where \( r \) is either 0 or \( N\left( r\right) < N\left( u\right) \) . In ei... | Yes |
Proposition 6. Let \( R \) be a Principal Ideal Domain and let \( a \) and \( b \) be nonzero elements of \( R \) . Let \( d \) be a generator for the principal ideal generated by \( a \) and \( b \) . Then\n\n(1) \( d \) is a greatest common divisor of \( a \) and \( b \)\n\n(2) \( d \) can be written as an \( R \) -l... | Proof: This is just Propositions 2 and 3. | No |
Proposition 7. Every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal. | Proof: Let \( \left( p\right) \) be a nonzero prime ideal in the Principal Ideal Domain \( R \) and let \( I = \left( m\right) \) be any ideal containing \( \left( p\right) \) . We must show that \( I = \left( p\right) \) or \( I = R \) . Now \( p \in \left( m\right) \) so \( p = {rm} \) for some \( r \in R \) . Since ... | Yes |
Corollary 8. If \( R \) is any commutative ring such that the polynomial ring \( R\left\lbrack x\right\rbrack \) is a Principal Ideal Domain (or a Euclidean Domain), then \( R \) is necessarily a field. | Proof: Assume \( R\left\lbrack x\right\rbrack \) is a Principal Ideal Domain. Since \( R \) is a subring of \( R\left\lbrack x\right\rbrack \) then \( R \) must be an integral domain (recall that \( R\left\lbrack x\right\rbrack \) has an identity if and only if \( R \) does). The ideal \( \left( x\right) \) is a nonzer... | Yes |
Proposition 9. The integral domain \( R \) is a P.I.D. if and only if \( R \) has a Dedekind-Hasse norm. | Proof: Let \( I \) be any nonzero ideal in \( R \) and let \( b \) be a nonzero element of \( I \) with \( N\left( b\right) \) minimal. Suppose \( a \) is any nonzero element in \( I \), so that the ideal \( \left( {a, b}\right) \) is contained in \( I \) . Then the Dedekind-Hasse condition on \( N \) and the minimalit... | No |
Proposition 10. In an integral domain a prime element is always irreducible. | Proof: Suppose \( \left( p\right) \) is a nonzero prime ideal and \( p = {ab} \) . Then \( {ab} = p \in \left( p\right) \), so by definition of prime ideal one of \( a \) or \( b \), say \( a \), is in \( \left( p\right) \) . Thus \( a = {pr} \) for some \( r \) . This implies \( p = {ab} = {prb} \) so \( {rb} = 1 \) a... | Yes |
Proposition 11. In a Principal Ideal Domain a nonzero element is a prime if and only if it is irreducible. | Proof: We have shown above that prime implies irreducible. We must show conversely that if \( p \) is irreducible, then \( p \) is a prime, i.e., the ideal \( \left( p\right) \) is a prime ideal. If \( M \) is any ideal containing \( \left( p\right) \) then by hypothesis \( M = \left( m\right) \) is a principal ideal. ... | Yes |
Proposition 12. In a Unique Factorization Domain a nonzero element is a prime if and only if it is irreducible. | Proof: Let \( R \) be a Unique Factorization Domain. Since by Proposition 10, primes of \( R \) are irreducible it remains to prove that each irreducible element is a prime. Let \( p \) be an irreducible in \( R \) and assume \( p \mid {ab} \) for some \( a, b \in R \) ; we must show that \( p \) divides either \( a \)... | Yes |
Proposition 13. Let \( a \) and \( b \) be two nonzero elements of the Unique Factorization Domain \( R \) and suppose\n\n\[ a = u{p}_{1}{}^{{e}_{1}}{p}_{2}{}^{{e}_{2}}\cdots {p}_{n}{}^{{e}_{n}}\;\text{ and }\;b = v{p}_{1}{}^{{f}_{1}}{p}_{2}{}^{{f}_{2}}\cdots {p}_{n}{}^{{f}_{n}} \]\n\nare prime factorizations for \( a ... | Proof: Since the exponents of each of the primes occurring in \( d \) are no larger than the exponents occurring in the factorizations of both \( a \) and \( b, d \) divides both \( a \) and \( b \) . To show that \( d \) is a greatest common divisor, let \( c \) be any common divisor of \( a \) and \( b \) and let \( ... | Yes |
Corollary 15. (Fundamental Theorem of Arithmetic) The integers \( \mathbb{Z} \) are a Unique Factorization Domain. | Proof: The integers \( \mathbb{Z} \) are a Euclidean Domain, hence are a Unique Factorization Domain by the theorem. | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.